Question

A student working in the laboratory produces 6.81 grams of calcium oxide, CaO, from 20.7 grams of calcium

Answer 1

Answer:

Percent yield = 57.8 %

Theoretical yield  = 11.781 g

Explanation:

Given data:

Mass of CaO produced = 6.81 g

Mass of CaCO₃ react = 20.7 g

Theoretical yield = ?

Percent yield = ?

Solution:

Chemical equation:

CaCO₃       →      CaO + CO₂

Number of moles of CaCO₃ :

Number of moles = mass/molar mass

Number of moles = 20.7 g/ 100.1 g/mol

Number of moles = 0.21 mol

Now we will compare the moles of CaCO₃ with CaO.

                  CaCO₃        :           CaO

                       1              :              1

                      0.21         :           0.21

Theoretical yield of CaO:  

Mass = number of moles × molar mass

Mass = 0.21 mol × 56.1 g/mol

Mass = 11.781 g

Percent yield:

Percent yield = ( actual yield / theoretical yield ) ×  100

Percent yield = (6.81 g/ 11.781 g) × 100

Percent yield = 57.8 %